algebra precalculus - How to explain what is wrong in this?

My friend showed this to me and I instantly know that this is wrong. However, I cannot explain why this is wrong to my friend.

Question. Prove $\displaystyle \frac{100-100}{100-100} = 2.$

Answer.
$$\begin{align*}
\frac{100-100}{100-100} &= \frac{(10)^2 - (10)^2}{10(10-10)}\\
&= \frac{(10+10)(10-10)}{10(10-10)}\\
&= \frac{20}{10} = 2.

\end{align*}$$

My argument is that in the third step, where it goes like this:
$$\frac{(10+10)(10-10)}{10(10-10)}$$
you cannot just cancel out the $(10-10)$ - it doesn't seem right. However, I am at a loss of explaining why exactly you cannot do that and my friend has the argumentative power (is that even a word? I mean he is good with arguments, even if they are not facts) and he has me confused to the point that I am starting to think it can be done.

Can anyone please explain why this is wrong?

Thanks.

Answer

There is no such thing as a "cancel" operation. This is, rather, shorthand for factoring out $1$, and simplifying. In other words, suppose you have
$$\frac{x^3+x}{x^2+1}$$
You could factor out
$$\frac{x^2+1}{x^2+1}$$
to yield
$$\frac{x(x^2+1)}{x^2+1}$$
However, since, everywhere
$$\frac{x^2+1}{x^2+1}=1$$
You simply replace the former with the latter, to yield

$$x.$$

At a fundamental level, where your proof goes wrong is that it skips over this subtlety and uses the "shortcut" of canceling without respect for the conditions under which that shortcut is valid.

In the end, what it comes down to is that
$$\frac{(10-10)}{(10-10)}=\frac{0}{0}\ne1,$$
which is the condition you need in order to make that cancellation.